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I gather that the impact speed depends on the radius of the orbit of the asteroid.

Is the orbit of asteroids in the same direction around the sun, or can they move in the "opposite" direction?

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Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other minor planets do move in opposite, or retrograde, orbits (see this Wikipedia article for a list of such objects).

The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e. $$ v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}. $$ Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.

The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is $$ v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1}, $$ where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).

If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to the escape velocity from the Sun at the location of Earth. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.

Hence, the maximum impact velocity is $$ v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}. $$

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  • $\begingroup$ The head on collision should be close to perpendicular which would require vector addition at close to a 90 degree angle. Also, you should add the 11 km/s for Earth's gravity. The highest velocity impact might work out to be pure retrograde not high eccentricity. $\endgroup$ – userLTK Apr 21 '17 at 22:07
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    $\begingroup$ @userLTK: Actually, it doesn't have to be perpendicular. You can think of it in reverse: If you launch an object tangentially to Earth's object, it will escape if the speed is ≥42 km/s (similarly, if you ride your bike along the bike lane at ≥11 km/s, you will leave Earth; you don't have to leave Earth perpendicularly). Also, I think that at a head-on collision, the increase in speed due to Earth's gravity can be neglected. $\endgroup$ – pela Apr 21 '17 at 22:31
  • $\begingroup$ Good points. I think you're right about that. Should I delete my comment? $\endgroup$ – userLTK Apr 21 '17 at 22:36
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    $\begingroup$ @userLTK: No, I think it's a good comment because you're definitely not the only one who thinks this. Intuitively, it seems easier to leave Earth "straight up" than at an angle, but that's not the case. Although in reality, the more inclined your path is, the more atmosphere you'd also have to go through. But if you leave Earth at an angle in the eastward direction, you gain some speed due to Earth's spin, so in practice I think they use some compromise between speed gain and not-too-much atmosphere (but this is spaceflight, of which I know next to nothing). $\endgroup$ – pela Apr 21 '17 at 22:51
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    $\begingroup$ @BSeven: First the disclaimer: I'm an theoretical astronomer, not a rocket scientist, so my knowledge about the real world is limited :) But I think that rockets are launched straight up, but then turn their path to an angle in the same direction as the rotation of Earth, benefitting from the rotation. Hmm… I wrote some more here, but I think I shouldn't, so perhaps go to Space Exploration SE instead. $\endgroup$ – pela Apr 24 '17 at 7:31

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